Complete Polynomial Factorization

Understanding Complete Factorization

We have learned to factor polynomials, for example using the Factor Theorem. However, sometimes the factorization result still leaves factors that are not linear (like quadratic factors) which cannot be factored further using real numbers.

Complete Factorization (or Complete Linear Factorization) is the process of factoring a polynomial into a product of linear factors, where these factors may involve complex numbers.

This concept is based on the Fundamental Theorem of Algebra, which states that every polynomial of degree $n \ge 1$ has exactly $n$ roots (zeros) in the set of complex numbers (including real roots and repeated roots).

Complete Factorization Property of Polynomials

If $P(x)$ is a polynomial of degree $n \ge 1$ with leading coefficient $a_n \neq 0$, then there exist complex numbers $c_1, c_2, \dots, c_n$ (which are the roots of $P(x)$) such that:

This means that every polynomial of degree $n$ can be broken down into exactly $n$ linear factors $(x - \text{root})$ multiplied by its leading coefficient.

Steps for Complete Factorization

To perform a complete factorization of a polynomial $P(x)$:

1. Find All Complex Roots: Find all $n$ complex roots (zeros) of $P(x) = 0$. This might involve:

   • Factoring directly (grouping, etc.).
   • Using the Rational Zero Theorem to find rational roots.
   • Using division (Horner/long division) to reduce the degree of the polynomial after a root is found.
   • Solving quadratic equations (using the quadratic formula) which might yield complex roots $a \pm bi$.

2. Apply the Factor Theorem: For each root $c_i$ found, form its linear factor, which is $(x - c_i)$.

3. Write the Complete Factorization: Multiply all the obtained linear factors by the leading coefficient $a_n$ of $P(x)$.

   $$P(x) = a_n(x - c_1)(x - c_2)\dots(x - c_n)$$

Using Complete Factorization

Find all complex zeros of $P(x) = x^3 - x^2 + x - 1$ and factor the polynomial completely.

Solution:

1. Find Roots: Let's try to factor $P(x)$ first.

   • Factor by grouping:

     $$P(x) = (x^3 - x^2) + (x - 1)$$

     $$P(x) = x^2(x - 1) + 1(x - 1)$$

     $$P(x) = (x^2 + 1)(x - 1)$$

   • Now, find the roots by setting $P(x) = 0$:

     $$(x^2 + 1)(x - 1) = 0$$

   • This gives two possibilities:

     • $x - 1 = 0 \implies x = 1$
     • $x^2 + 1 = 0 \implies x^2 = -1 \implies x = \pm\sqrt{-1} \implies x = \pm i$

   So, the complex roots are $1, i, -i$.

2. Form Linear Factors:

   • From the root $1$, the factor is $(x - 1)$.
   • From the root $i$, the factor is $(x - i)$.
   • From the root $-i$, the factor is $(x - (-i)) = (x + i)$.

3. Write Complete Factorization:

   The leading coefficient of $P(x)$ is $a_3 = 1$.

   $$P(x) = 1 \cdot (x - 1)(x - i)(x + i)$$

   $$P(x) = (x - 1)(x - i)(x + i)$$

Exercise

Find all complex zeros of $P(x) = x^3 - 3x^2 + x + 5$, then factor $P(x)$ completely.

Answer Key

1. Find Rational Roots (Rational Zero Theorem):

   • $a_0 = 5$, factors $p$: $\pm 1, \pm 5$.

   • $a_n = 1$, factors $q$: $\pm 1$.

   • Possible roots $p/q$: $\pm 1, \pm 5$.

   • Test $x = -1$:

     $$P(-1) = (-1)^3 - 3(-1)^2 + (-1) + 5 = -1 - 3(1) - 1 + 5 = -1 - 3 - 1 + 5 = 0$$

     So, $x = -1$ is a root, and $(x+1)$ is a factor.

2. Divide using Horner's Method ($c = -1$):

   $$\begin{array}{c|cccc} -1 & 1 & -3 & 1 & 5 \\ & & -1 & 4 & -5 \\ \hline & 1 & -4 & 5 & \boxed{0} \\ \end{array}$$

   Quotient $H(x) = x^2 - 4x + 5$.

   $P(x) = (x+1)(x^2 - 4x + 5)$.

3. Find Roots of the Quotient: Solve $x^2 - 4x + 5 = 0$ using the quadratic formula.

   $$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)}$$

   $$x = \frac{4 \pm \sqrt{16 - 20}}{2}$$

   $$x = \frac{4 \pm \sqrt{-4}}{2}$$

   $$x = \frac{4 \pm 2i}{2}$$

   $$x = 2 \pm i$$

   The other roots are $2 + i$ and $2 - i$.

4. All Complex Roots: The roots are $-1, 2+i, 2-i$.

5. Complete Factorization ($a_n = 1$):

   $$P(x) = 1 \cdot (x - (-1))(x - (2+i))(x - (2-i))$$

   $$P(x) = (x + 1)(x - 2 - i)(x - 2 + i)$$

Complex Conjugate Root Pairs

An important question arises: Is it possible for a polynomial with all real coefficients and constant term to have exactly one complex zero that is not a real number?

The answer is impossible.

This is due to the property of complex conjugate pairs. If a polynomial has real coefficients, then its non-real complex roots ($a + bi$ with $b \neq 0$) always occur in conjugate pairs.

This means that if $a + bi$ is a root, then its conjugate, $a - bi$, must also be a root of the polynomial.

Therefore, non-real complex roots cannot appear alone; they always come in pairs. Thus, it is impossible to have exactly one non-real complex root for a polynomial with real coefficients.